\(\int \frac {c+d x^4}{x^2 (a+b x^4)^{13/4}} \, dx\) [175]

Optimal result

Integrand size = 22, antiderivative size = 151 \[ \int \frac {c+d x^4}{x^2 \left (a+b x^4\right )^{13/4}} \, dx=-\frac {c}{a x \left (a+b x^4\right )^{9/4}}-\frac {(10 b c-a d) x^3}{9 a^2 \left (a+b x^4\right )^{9/4}}-\frac {2 (10 b c-a d) x^3}{15 a^3 \left (a+b x^4\right )^{5/4}}+\frac {4 (10 b c-a d) \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{15 a^{7/2} \sqrt {b} \sqrt [4]{a+b x^4}} \] Output:

-c/a/x/(b*x^4+a)^(9/4)-1/9*(-a*d+10*b*c)*x^3/a^2/(b*x^4+a)^(9/4)-2/15*(-a* 
d+10*b*c)*x^3/a^3/(b*x^4+a)^(5/4)+4/15*(-a*d+10*b*c)*(1+a/b/x^4)^(1/4)*x*E 
llipticE(sin(1/2*arccot(b^(1/2)*x^2/a^(1/2))),2^(1/2))/a^(7/2)/b^(1/2)/(b* 
x^4+a)^(1/4)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.03 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.56 \[ \int \frac {c+d x^4}{x^2 \left (a+b x^4\right )^{13/4}} \, dx=-\frac {c}{a x \left (a+b x^4\right )^{9/4}}-\frac {(10 b c-a d) x^3 \sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {13}{4},\frac {7}{4},-\frac {b x^4}{a}\right )}{3 a^4 \sqrt [4]{a+b x^4}} \] Input:

Integrate[(c + d*x^4)/(x^2*(a + b*x^4)^(13/4)),x]
 

Output:

-(c/(a*x*(a + b*x^4)^(9/4))) - ((10*b*c - a*d)*x^3*(1 + (b*x^4)/a)^(1/4)*H 
ypergeometric2F1[3/4, 13/4, 7/4, -((b*x^4)/a)])/(3*a^4*(a + b*x^4)^(1/4))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {955, 819, 819, 813, 858, 807, 212}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {c+d x^4}{x^2 \left (a+b x^4\right )^{13/4}} \, dx\)

\(\Big \downarrow \) 955

\(\displaystyle -\frac {(10 b c-a d) \int \frac {x^2}{\left (b x^4+a\right )^{13/4}}dx}{a}-\frac {c}{a x \left (a+b x^4\right )^{9/4}}\)

\(\Big \downarrow \) 819

\(\displaystyle -\frac {(10 b c-a d) \left (\frac {2 \int \frac {x^2}{\left (b x^4+a\right )^{9/4}}dx}{3 a}+\frac {x^3}{9 a \left (a+b x^4\right )^{9/4}}\right )}{a}-\frac {c}{a x \left (a+b x^4\right )^{9/4}}\)

\(\Big \downarrow \) 819

\(\displaystyle -\frac {(10 b c-a d) \left (\frac {2 \left (\frac {2 \int \frac {x^2}{\left (b x^4+a\right )^{5/4}}dx}{5 a}+\frac {x^3}{5 a \left (a+b x^4\right )^{5/4}}\right )}{3 a}+\frac {x^3}{9 a \left (a+b x^4\right )^{9/4}}\right )}{a}-\frac {c}{a x \left (a+b x^4\right )^{9/4}}\)

\(\Big \downarrow \) 813

\(\displaystyle -\frac {(10 b c-a d) \left (\frac {2 \left (\frac {2 x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{5/4} x^3}dx}{5 a b \sqrt [4]{a+b x^4}}+\frac {x^3}{5 a \left (a+b x^4\right )^{5/4}}\right )}{3 a}+\frac {x^3}{9 a \left (a+b x^4\right )^{9/4}}\right )}{a}-\frac {c}{a x \left (a+b x^4\right )^{9/4}}\)

\(\Big \downarrow \) 858

\(\displaystyle -\frac {(10 b c-a d) \left (\frac {2 \left (\frac {x^3}{5 a \left (a+b x^4\right )^{5/4}}-\frac {2 x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{5/4} x}d\frac {1}{x}}{5 a b \sqrt [4]{a+b x^4}}\right )}{3 a}+\frac {x^3}{9 a \left (a+b x^4\right )^{9/4}}\right )}{a}-\frac {c}{a x \left (a+b x^4\right )^{9/4}}\)

\(\Big \downarrow \) 807

\(\displaystyle -\frac {(10 b c-a d) \left (\frac {2 \left (\frac {x^3}{5 a \left (a+b x^4\right )^{5/4}}-\frac {x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4}}d\frac {1}{x^2}}{5 a b \sqrt [4]{a+b x^4}}\right )}{3 a}+\frac {x^3}{9 a \left (a+b x^4\right )^{9/4}}\right )}{a}-\frac {c}{a x \left (a+b x^4\right )^{9/4}}\)

\(\Big \downarrow \) 212

\(\displaystyle -\frac {(10 b c-a d) \left (\frac {2 \left (\frac {x^3}{5 a \left (a+b x^4\right )^{5/4}}-\frac {2 x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right )\right |2\right )}{5 a^{3/2} \sqrt {b} \sqrt [4]{a+b x^4}}\right )}{3 a}+\frac {x^3}{9 a \left (a+b x^4\right )^{9/4}}\right )}{a}-\frac {c}{a x \left (a+b x^4\right )^{9/4}}\)

Input:

Int[(c + d*x^4)/(x^2*(a + b*x^4)^(13/4)),x]
 

Output:

-(c/(a*x*(a + b*x^4)^(9/4))) - ((10*b*c - a*d)*(x^3/(9*a*(a + b*x^4)^(9/4) 
) + (2*(x^3/(5*a*(a + b*x^4)^(5/4)) - (2*(1 + a/(b*x^4))^(1/4)*x*EllipticE 
[ArcTan[Sqrt[a]/(Sqrt[b]*x^2)]/2, 2])/(5*a^(3/2)*Sqrt[b]*(a + b*x^4)^(1/4) 
)))/(3*a)))/a
 

Defintions of rubi rules used

rule 212

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) 
)*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 813

Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Simp[x*((1 + a/(b*x^4) 
)^(1/4)/(b*(a + b*x^4)^(1/4)))   Int[1/(x^3*(1 + a/(b*x^4))^(5/4)), x], x] 
/; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 819

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-( 
c*x)^(m + 1))*((a + b*x^n)^(p + 1)/(a*c*n*(p + 1))), x] + Simp[(m + n*(p + 
1) + 1)/(a*n*(p + 1))   Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a 
, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p 
, x]
 

rule 858

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

rule 955

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), 
 x] + Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1))   Int[(e 
*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b* 
c - a*d, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || 
(LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]
 

Maple [F]

\[\int \frac {d \,x^{4}+c}{x^{2} \left (b \,x^{4}+a \right )^{\frac {13}{4}}}d x\]

Input:

int((d*x^4+c)/x^2/(b*x^4+a)^(13/4),x)
 

Output:

int((d*x^4+c)/x^2/(b*x^4+a)^(13/4),x)
 

Fricas [F]

\[ \int \frac {c+d x^4}{x^2 \left (a+b x^4\right )^{13/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {13}{4}} x^{2}} \,d x } \] Input:

integrate((d*x^4+c)/x^2/(b*x^4+a)^(13/4),x, algorithm="fricas")
 

Output:

integral((b*x^4 + a)^(3/4)*(d*x^4 + c)/(b^4*x^18 + 4*a*b^3*x^14 + 6*a^2*b^ 
2*x^10 + 4*a^3*b*x^6 + a^4*x^2), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 110.50 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.54 \[ \int \frac {c+d x^4}{x^2 \left (a+b x^4\right )^{13/4}} \, dx=\frac {c \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {13}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {13}{4}} x \Gamma \left (\frac {3}{4}\right )} + \frac {d x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {13}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {13}{4}} \Gamma \left (\frac {7}{4}\right )} \] Input:

integrate((d*x**4+c)/x**2/(b*x**4+a)**(13/4),x)
 

Output:

c*gamma(-1/4)*hyper((-1/4, 13/4), (3/4,), b*x**4*exp_polar(I*pi)/a)/(4*a** 
(13/4)*x*gamma(3/4)) + d*x**3*gamma(3/4)*hyper((3/4, 13/4), (7/4,), b*x**4 
*exp_polar(I*pi)/a)/(4*a**(13/4)*gamma(7/4))
 

Maxima [F]

\[ \int \frac {c+d x^4}{x^2 \left (a+b x^4\right )^{13/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {13}{4}} x^{2}} \,d x } \] Input:

integrate((d*x^4+c)/x^2/(b*x^4+a)^(13/4),x, algorithm="maxima")
 

Output:

integrate((d*x^4 + c)/((b*x^4 + a)^(13/4)*x^2), x)
 

Giac [F]

\[ \int \frac {c+d x^4}{x^2 \left (a+b x^4\right )^{13/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {13}{4}} x^{2}} \,d x } \] Input:

integrate((d*x^4+c)/x^2/(b*x^4+a)^(13/4),x, algorithm="giac")
 

Output:

integrate((d*x^4 + c)/((b*x^4 + a)^(13/4)*x^2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {c+d x^4}{x^2 \left (a+b x^4\right )^{13/4}} \, dx=\int \frac {d\,x^4+c}{x^2\,{\left (b\,x^4+a\right )}^{13/4}} \,d x \] Input:

int((c + d*x^4)/(x^2*(a + b*x^4)^(13/4)),x)
 

Output:

int((c + d*x^4)/(x^2*(a + b*x^4)^(13/4)), x)
 

Reduce [F]

\[ \int \frac {c+d x^4}{x^2 \left (a+b x^4\right )^{13/4}} \, dx=\left (\int \frac {x^{2}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b \,x^{4}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{2} x^{8}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{3} x^{12}}d x \right ) d +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3} x^{2}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b \,x^{6}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{2} x^{10}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{3} x^{14}}d x \right ) c \] Input:

int((d*x^4+c)/x^2/(b*x^4+a)^(13/4),x)
 

Output:

int(x**2/((a + b*x**4)**(1/4)*a**3 + 3*(a + b*x**4)**(1/4)*a**2*b*x**4 + 3 
*(a + b*x**4)**(1/4)*a*b**2*x**8 + (a + b*x**4)**(1/4)*b**3*x**12),x)*d + 
int(1/((a + b*x**4)**(1/4)*a**3*x**2 + 3*(a + b*x**4)**(1/4)*a**2*b*x**6 + 
 3*(a + b*x**4)**(1/4)*a*b**2*x**10 + (a + b*x**4)**(1/4)*b**3*x**14),x)*c